mysql登錄用戶名密碼

Question

$servername = "localhost"; 
$username = "csc4370FA14_18"; 
$password = "1db23"; 
$dbname = "csc4370FA14_18"; 


$conn = mysqli_connect($servername, $username, $password, $dbname); 

if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
$username_login = $_POST["username"]; 
$password_login = $_POST["pw"]; 

$query2 = mysql_query("SELECT * FROM users WHERE name='$username_login'"); 
$numrow = mysql_num_rows($query2); 

if ($numrow != 0) { 
    while ($row = mysql_fetch_assoc($query2)) { 
     $dbusername = $row['name']; 
     $dbpassword = $row['password']; 
    } 

    // Check to see if username and password match 
    if ($username_login==$dbusername && $password_login==$dbpassword) { 
     echo "You are in"; 
    } 
    else { 
     echo "Sorry $username_login. Incorrect password!"; 
    } 
} 

這是我用來檢查用戶是否匹配表中的密碼（同一行）的代碼。 我得到的錯誤：

警告：mysql_query（）：拒絕用戶'apache'@'localhost'（使用密碼：否）/ home/csc4370FA14_18/public_html/program/assignments/group project3/login第14行警告：mysql_query（）：無法在第14行的/ home/csc4370FA14_18/public_html/program/assignments/group project3/login.php中建立到服務器的鏈接警告：mysql_num_rows（）期望參數1到是資源，在/ home/csc4370fa14_18/public_html/program/assignments/group project3/login.php在線15給出的布爾值

我不知道爲什麼這可能是不正確的，因爲登錄憑證等工作正常。我認爲這與mysqli有關，但我對mysql_ *函數的掌握不夠。請幫忙！我知道這是一個正確的連接信息。

Answer 1

試試看看這個代碼。

$servername = "localhost"; 
$username = "csc4370FA14_18"; 
$password = "1db23"; 
$dbname = "csc4370FA14_18"; 


$conn = mysqli_connect($servername, $username, $password, $dbname); 

if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
$username_login = $_POST["username"]; 
$password_login = $_POST["pw"]; 

$query2 = mysqli_query($conn,"SELECT * FROM users WHERE name='$username_login'"); 
$numrow = mysqli_num_rows($query2); 

if ($numrow != 0) { 
    while ($row = mysqli_fetch_assoc($query2)) { 
     $dbusername = $row['name']; 
     $dbpassword = $row['password']; 
    } 

    // Check to see if username and password match 
    if ($username_login==$dbusername && $password_login==$dbpassword) { 
     echo "You are in"; 
    } 
    else { 
     echo "Sorry $username_login. Incorrect password!"; 
    } 
}